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Is there a positive decreasing sequence $(a_n)_{n\ge 0}$ such that

${\it i.}$ $\lim\limits_{n\to\infty} a_n=0$.

${\it ii.}$ $\lim\limits_{n\to\infty} na_n=\infty$.

${\it iii.}$ there is a real $\ell$ such that $\lim\limits_{n\to\infty} \left(na_{n+1}-\sum_{k=1}^n a_k\right)=\ell.$

Of course without condition ${\it i.}$ constant sequences are non-increasing examples that satisfy the other two conditions, but the requirement that the sequence must be decreasing to zero makes it hard to find!.

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  • $\begingroup$ Interesting question! Is there some context to it, or was it just invented as a challenge problem? $\endgroup$ – rubik Nov 30 '15 at 15:51
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    $\begingroup$ Without iii. it's also straightforward: $a_n = 1/\sqrt n$ for $n>0$, $a_0 =$ anything. $\endgroup$ – BrianO Nov 30 '15 at 15:56
  • $\begingroup$ Even $l$ is not relevant, it can be about anything. If $l$ is a solution, any (positive) multiple of $l$ is also a solution, just multiply all $a_i$. But the limit has to exist. $\endgroup$ – Pieter21 Nov 30 '15 at 16:01
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    $\begingroup$ @rubik, well, in fact I was preparing a homework for my students, then I discovered that it was more a challenge for myself than it was for them. $\endgroup$ – Omran Kouba Nov 30 '15 at 16:08
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    $\begingroup$ @rubik I think the sequence in ${\it iii}$ is negative decreasing, so $\ell$ can only be negative. $\endgroup$ – Omran Kouba Nov 30 '15 at 17:02
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There is no such sequence. Let $s_n=\sum_{k=1}^n a_k$. By (iii) we can write $$ n(s_{n+1}-s_n) - s_n = \ell + b_n = \ell(n+1-n)+ b_n$$ where $b_n\to0$. Then it follows that for all $n\ge1$, $$ \frac{s_{n+1}+\ell}{n+1} = \frac{s_n+\ell}n + \frac{b_n}{n(n+1)} = \frac{s_1+\ell}1 + \sum_{k=1}^n \frac{b_k}{k(k+1)} = A+c_{n+1},$$ where $$ A = s_1+\ell+\sum_{k=1}^\infty \frac{b_k}{k(k+1)}$$ is finite, and $$ c_{n+1} = -\sum_{k=n+1}^\infty \frac{b_k}{k(k+1)},$$ which satisfies $$ |c_n| \le \sup_{k\ge n}|b_k| \sum_{k=n}^\infty \left( \frac1k-\frac{1}{k+1}\right) = \frac1n \sup_{k\ge n}|b_k| $$ Hence $nc_n\to 0$ as $n\to\infty$.

But now it follows $$a_{n+1} = (s_{n+1}+\ell)-(s_n+\ell) = (n+1)(A+c_{n+1})-n(A+c_n) \to A,$$ hence $A=0$ by (i). Then it follows further that $$ s_n + \ell = n c_n \to 0,$$ whence the series $\sum_{k=1}^\infty a_k$ converges to $-\ell$. But this is impossible, since by (ii) we have $a_k\ge1/k$ for $k$ large enough.

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A try:

Suppose that such a sequence exist. Put $\displaystyle na_{n+1}=\sum_{k=1}^n a_k+L+e_n$, with $e_n\to 0$. Replacing $n$ by $n+1$, we get $$(n+1)(a_{n+1}-a_{n+2})=e_n-e_{n+1}$$

As $a_n$ is decreasing, we get that $e_n$ is decreasing too. Now as $e_n\to 0$, we have $e_n\geq 0$, and $\displaystyle a_{n+1}-a_{n+2}=\frac{e_n-e_{n+1}}{n+1} \leq \frac{e_n}{n}-\frac{e_{n+1}}{n+1}$. This show that for $m\geq 2$ we have $$a_{n+1}\leq a_{n+m+2}+\frac{e_n}{n}-\frac{e_{n+m+1}}{n+m+1}\leq a_{n+m+1}+\frac{e_n}{n}$$ If $m\to +\infty$, we get $na_{n+1}\leq e_n$, hence $na_n\to 0$, and condition ii) is not satisfied when conditions (i) and (iii) are.

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Assume $\{a_n\}$ exists.

Let $a_n = f (n)$. By i and ii, we have $f (n) = o (1)$ and $f (n) = \Omega (1/n)$. By Euler-McLaurin summation formula, we have $$\sum_{k = 1}^{n} f (k) = c_1 F (n) + o (F (n))$$ for a constant $c_1$, where $F$ is anti-derivative of $f$, and $F (n) = o (n)$ and $F (n) = \Omega (1)$. Then, by iii, we have

$\begin {eqnarray} n a_{n + 1} - \sum_{k = 1}^{n} a_k & = & n f (n + 1) - \left(c_1 F (n) + o (F (n)\right) \nonumber \\ & = & c_2 F (n) + o (F (n)) - \left(c_1 F (n) + o (F (n))\right) \nonumber \\ & = & (c_1 - c_2) F (n) + o (F (n)) \nonumber \\ & = & O (F (n)) \nonumber \\ & = & \Omega (1), \end {eqnarray}$

but $\Omega (1)_{n \to \infty} = \infty$, a contradiction. Hence, $\{a_n\}$ does not exist.

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  • $\begingroup$ Why we cannot have $a_n=1/(ln n)$ for example! wrong answer. $\endgroup$ – Omran Kouba Dec 1 '15 at 16:53
  • $\begingroup$ Because $a_1$ would be undefined! Wrong counterexample. $\endgroup$ – user98186 Dec 1 '15 at 16:57
  • $\begingroup$ Well take $a_n=1/(\ln(n+217))$. $\endgroup$ – Omran Kouba Dec 1 '15 at 18:13
  • $\begingroup$ OK... but the thing is you didn't get my idea, you right away jumped to bashing me. I took $a_n = n^c$ but I could take $a_n = M n^c$ for a constant $M$ as well. The idea here is $a_n = o (n)$ but $\Omega (1)$, and my choice $a_n = n^c$ represents this extremely well. In fact, take $a_n = f (n)$ for any $f(n) = o (n)$ and $\Omega (1)$ Bye. $\endgroup$ – user98186 Dec 1 '15 at 19:07
  • $\begingroup$ I think that if $a_n=Mn^c$ with $0<c<1$ then $(a_n)$ cannot be decreasing to $0$, and the fact that somthing is $o(n)$ or $\Omega(1)$ does not make it monotonous,.....OK, Bye. $\endgroup$ – Omran Kouba Dec 1 '15 at 20:48

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